Question

A cookie jar is moving up a 40° incline. At a point 55 cm from the bottom of the incline (measured along the incline), the jar has a speed of 1.8 m/s. The coefficient of kinetic friction between jar and incline is 0.15.

(a) How much farther up the incline will the jar move?
___ m

(b) How fast will it be going when it has slid back to the bottom of the incline?
___m/s

Answer 1

a)

from the force diagram ,

F_n = nomal force = mgCos40

frictional force is given as

f = _k Fn = _k mgCos40

Net force opposite to motion of jar is given as

F_net = f + mgSin40

F_net = _k mgCos40 + mgSin40

But F_net = ma

so ma = _k mgCos40 + mgSin40

a = (0.15) (9.8) Cos40 + (9.8) Sin40

a = 7.43 m/s²

a)

V_i = initial velocity = 1.8 m/s

V_f = final velocity = 0 m/s

a = - 7.43 m/s²

d = stopping distance

Using the equation

V_f² = V_i² + 2 a d

0² = 1.8² + 2 (- 7.43) d

d = 0.218 m = 21.8 cm

b)

v = speed at bottom

Total length of the incline , L = 55 + 21.8 = 76.8 cm = 0.768 m

from triangle ABC :

Sin40 = BC/AB = BC /L

height of incline = h = L Sin40 = 0.768 Sin40 = 0.494 m

Using conservation of energy

Potential energy at Top = Kinetic energy at Bottom + work done by friction

mgh = (0.5) m v² + fL

mgh = (0.5) m v² + (_k mgCos40)L

2gh = v² + (2 _k gCos40)L

2(9.8) (0.494) = v² + (2 (0.15) (9.8)Cos40) (0.768)

V = 2.82 m/s